Given the overall formation constant of the $[Fe(CN)_6]^{-4}$ ion as $10^{35}$ and the standard potential for half reaction $Fe^{+3}+e^-\leftrightharpoons Fe^{+2},E^0=0.77V$,$[Fe(CN)_6]^{-3}+e^-\leftrightharpoons [Fe(CN)_6]^{-4},E^0=0.36V$.Then the overall formation constant of $[Fe(CN)_6]^{-3}$ ion is

Question

$\begin{array}{1 1}(a)\;8\times 10^{92}\\(b)\;6.3\times 10^{14}\\(c)\;8.59\times 10^{41}\\(d)\;9\times 10^{36}\end{array}$